TITLE: image of adjoint equals orthogonal complement of kernel
QUESTION [18 upvotes]: Let $T:V\to W$ be a linear map of finite-dimensional spaces.  Then
$${\rm im}(T^{\textstyle*})=({\rm ker}\,T)^\perp\ .\tag{$*$}$$
I can prove this as follows:
$${\rm ker}(T^{\textstyle*})=({\rm im}\,T)^\perp\tag{$**$}$$
is quite easy, and we know $T^{\textstyle*}{}^{\textstyle*}=T$ and $W^{\perp\perp}=W$, so
$${\rm im}(T^{\textstyle*})=({\rm im}\,T^{\textstyle*})^{\perp\perp}=({\rm ker}\,T^{\textstyle*}{}^{\textstyle*})^\perp=({\rm ker}\,T)^\perp\ .$$
However, I would be interested in a "direct" proof of $(*)$.  It's fairly easy to show ${\rm LHS}\subseteq{\rm RHS}$.  For the converse I have tried obvious things but seem to be going round in circles.
Also, any insights as to why $(**)$ is harder than $(*)$ - if in fact it is :)
Edit.  To clarify, I am considering the adjoint defined in terms of an inner product,
$$\langle\,T({\bf v})\mid{\bf w}\,\rangle
  =\langle\,{\bf v}\mid T^{\textstyle*}({\bf w})\,\rangle\ .$$

REPLY [12 votes]: Let ${\bf v}\in{\rm im}(T^\ast)$, then ${\bf v}=T^\ast({\bf w})$ for some ${\bf w}\in W$. Now, given ${\bf u}\in\ker T$, we see that $T({\bf u})={\bf 0}$ and therefore
$$\langle {\bf u}\mid{\bf v}\rangle
=\langle {\bf u}\mid T^\ast({\bf w})\rangle
=\langle T({\bf u})\mid {\bf w}\rangle
=\langle {\bf 0}\mid{\bf w}\rangle
=0.$$
That is, ${\bf v}\in(\ker T)^\perp$. Conversely, if ${\bf v}\notin{\rm im}(T^\ast)$, then
there exists an ${\bf v}'\in{\rm im}(T^\ast)^\perp$ such that $\langle {\bf v}\mid{\bf v}'\rangle\ne 0$.
In fact, we have ${\bf v}'\in\ker T$ because $T^\ast T({\bf v}')\in{\rm im}(T^\ast)$, which implies
$$\langle T({\bf v}')\mid T({\bf v}')\rangle
=\langle {\bf v}'\mid T^\ast T({\bf v}')\rangle=0\quad\Longrightarrow\quad
T({\bf v}')={\bf 0}.$$
Therefore ${\bf v}\notin(\ker T)^\perp$, which completes the proof.